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## 7. The roots of each of the following quadratic equations are real and equal, find k. (1) 3$y^2$ + ky + l2 = 0 (2) kx (x – 2) + 6 = 0

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(1)    3$y^2$ + ky + l2 = 0
Sol.  3$y^2$ + ky + l2 = 0
Comparing with a$x^2$ + bx + c = 0
we get, a = 3, b = k and c = 12

Let’s find value of discriminant
Δ   =   D  =   $b^2$ – 4ac
=   $k^2$ – 4 × 3 × 12
=   $k^2$ – 144                               ...(I)
but roots of equation are real and equal
∴     Δ  =   D    =   0                        ...(II)
∴     $k^2$ – 144  =   0               [From (I) & (II)]
∴     $k^2$ – 122  =   0
By using $a^2$$b^2$ = (a + b) (a – b)
∴   (k + 12)  (k – 12)  =  0
∴   k + 12  =  0    or   k – 12  =  0
∴          k   = –12 or          k   = 12

(2)   kx (x – 2) + 6 = 0
Sol. kx (x – 2) + 6 = 0
∴     k$x^2$ – 2kx + 6 = 0
comparing with a$x^2$ + bx + c = 0
∴    a = k, b = –2k and c = 6
Let’s find value of discriminant
Δ  =  D  =  $b^2$ – 4ac
=  $(–2k)^2$ – 4 × k × 6
=  4$k^2$ – 24k                             ...(I)
But roots of equation are real and equal
∴    Δ   =   D     =  0                                 ...(II)
∴    4$k^2$ – 24k  =  0                    [From (I) & (II)]
∴    4k(k – 6)    =  0
k – 6    =  $0\over 4k$
∴           k – 6    =  0
∴                 k    =  6

uÉaÉïxÉqÉÏMüUhÉ August 10 , 2018 0 Comments 90 views